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A compression spring is made of music wire of 2 mm diameter having a shear strength and shear modulus of 800 MPa and 80 GPa respectively. The mean coil diameter is 20 mm, free length is 40 mm and the number of active coils is 10. If the mean coil diameter is reduced to 10 mm, the stiffness of the spring is approximately
- a)Decreased by 8 times
- b)Decreased by 2 times
- c)Increased by 2 times
- d)Increased by 8 times
Correct answer is option 'D'. Can you explain this answer?
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Compression springs are mechanical devices used to store and release energy. They are commonly used in various applications such as automotive suspensions, industrial machinery, and household appliances. The stiffness of a compression spring determines its ability to resist deformation under an applied load.
Given data:
- Diameter of music wire (d): 2 mm
- Shear strength (τ): 800 MPa
- Shear modulus (G): 80 GPa
- Mean coil diameter (Dm): 20 mm
- Free length (L0): 40 mm
- Number of active coils (Na): 10
To determine the stiffness of the spring, we need to calculate the spring constant (k). The spring constant is a measure of the force required to compress or extend a spring by a certain distance.
1. Calculate the shear stress:
Shear stress (τ) can be calculated using the formula:
τ = (16 * Force * Dm) / (π * d^3 * Na)
2. Calculate the shear modulus:
Shear modulus (G) is given in the question as 80 GPa.
3. Calculate the spring constant:
The spring constant (k) can be calculated using the formula:
k = (G * d^4 * Na) / (8 * Dm^3)
Now, let's calculate the initial stiffness of the spring using the given data.
Initial spring constant (k1) = (80 * 10^9 * (2 * 10^-3)^4 * 10) / (8 * (20 * 10^-3)^3) = 0.16 N/mm
Now, let's calculate the final stiffness of the spring when the mean coil diameter is reduced to 10 mm.
Final mean coil diameter (Dm') = 10 mm
Final spring constant (k2) = (80 * 10^9 * (2 * 10^-3)^4 * 10) / (8 * (10 * 10^-3)^3) = 1.28 N/mm
The ratio of the final spring constant to the initial spring constant is:
k2 / k1 = 1.28 / 0.16 = 8
Therefore, the stiffness of the spring is increased by 8 times when the mean coil diameter is reduced to 10 mm. The correct answer is option D.
Given data:
- Diameter of music wire (d): 2 mm
- Shear strength (τ): 800 MPa
- Shear modulus (G): 80 GPa
- Mean coil diameter (Dm): 20 mm
- Free length (L0): 40 mm
- Number of active coils (Na): 10
To determine the stiffness of the spring, we need to calculate the spring constant (k). The spring constant is a measure of the force required to compress or extend a spring by a certain distance.
1. Calculate the shear stress:
Shear stress (τ) can be calculated using the formula:
τ = (16 * Force * Dm) / (π * d^3 * Na)
2. Calculate the shear modulus:
Shear modulus (G) is given in the question as 80 GPa.
3. Calculate the spring constant:
The spring constant (k) can be calculated using the formula:
k = (G * d^4 * Na) / (8 * Dm^3)
Now, let's calculate the initial stiffness of the spring using the given data.
Initial spring constant (k1) = (80 * 10^9 * (2 * 10^-3)^4 * 10) / (8 * (20 * 10^-3)^3) = 0.16 N/mm
Now, let's calculate the final stiffness of the spring when the mean coil diameter is reduced to 10 mm.
Final mean coil diameter (Dm') = 10 mm
Final spring constant (k2) = (80 * 10^9 * (2 * 10^-3)^4 * 10) / (8 * (10 * 10^-3)^3) = 1.28 N/mm
The ratio of the final spring constant to the initial spring constant is:
k2 / k1 = 1.28 / 0.16 = 8
Therefore, the stiffness of the spring is increased by 8 times when the mean coil diameter is reduced to 10 mm. The correct answer is option D.
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A compression spring is made of music wire of 2 mm diameter having a shear strength and shear modulus of 800 MPa and 80 GPa respectively. The mean coil diameter is 20 mm, free length is 40 mm and the number of active coils is 10. If the mean coil diameter is reduced to 10 mm, the stiffness of the spring is approximatelya)Decreased by 8 timesb)Decreased by 2 timesc)Increased by 2 timesd)Increased by 8 timesCorrect answer is option 'D'. Can you explain this answer?
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A compression spring is made of music wire of 2 mm diameter having a shear strength and shear modulus of 800 MPa and 80 GPa respectively. The mean coil diameter is 20 mm, free length is 40 mm and the number of active coils is 10. If the mean coil diameter is reduced to 10 mm, the stiffness of the spring is approximatelya)Decreased by 8 timesb)Decreased by 2 timesc)Increased by 2 timesd)Increased by 8 timesCorrect answer is option 'D'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A compression spring is made of music wire of 2 mm diameter having a shear strength and shear modulus of 800 MPa and 80 GPa respectively. The mean coil diameter is 20 mm, free length is 40 mm and the number of active coils is 10. If the mean coil diameter is reduced to 10 mm, the stiffness of the spring is approximatelya)Decreased by 8 timesb)Decreased by 2 timesc)Increased by 2 timesd)Increased by 8 timesCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A compression spring is made of music wire of 2 mm diameter having a shear strength and shear modulus of 800 MPa and 80 GPa respectively. The mean coil diameter is 20 mm, free length is 40 mm and the number of active coils is 10. If the mean coil diameter is reduced to 10 mm, the stiffness of the spring is approximatelya)Decreased by 8 timesb)Decreased by 2 timesc)Increased by 2 timesd)Increased by 8 timesCorrect answer is option 'D'. Can you explain this answer?.
A compression spring is made of music wire of 2 mm diameter having a shear strength and shear modulus of 800 MPa and 80 GPa respectively. The mean coil diameter is 20 mm, free length is 40 mm and the number of active coils is 10. If the mean coil diameter is reduced to 10 mm, the stiffness of the spring is approximatelya)Decreased by 8 timesb)Decreased by 2 timesc)Increased by 2 timesd)Increased by 8 timesCorrect answer is option 'D'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A compression spring is made of music wire of 2 mm diameter having a shear strength and shear modulus of 800 MPa and 80 GPa respectively. The mean coil diameter is 20 mm, free length is 40 mm and the number of active coils is 10. If the mean coil diameter is reduced to 10 mm, the stiffness of the spring is approximatelya)Decreased by 8 timesb)Decreased by 2 timesc)Increased by 2 timesd)Increased by 8 timesCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A compression spring is made of music wire of 2 mm diameter having a shear strength and shear modulus of 800 MPa and 80 GPa respectively. The mean coil diameter is 20 mm, free length is 40 mm and the number of active coils is 10. If the mean coil diameter is reduced to 10 mm, the stiffness of the spring is approximatelya)Decreased by 8 timesb)Decreased by 2 timesc)Increased by 2 timesd)Increased by 8 timesCorrect answer is option 'D'. Can you explain this answer?.
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